Question-213217

Question Number 213217 by Spillover last updated on 01/Nov/24

Answered by MrGaster last updated on 01/Nov/24

Σ_(n=1) ^∞ (( (((2n)),(n) )+n)/(4^n (2n+1)^2 ))=Σ_(n=1) ^∞ ( (((2n)),(n) )/(4^n (2n+1)^2 ))+Σ_(n=1) ^∞ (n/(4^n (2n+1)^2 )) =Σ_(n=1) ^∞ ( (((2n)),(n) )/(4^n (2n+1)^2 ))+Σ_(n=1) ^∞ (n/(4^n (2n+1)^2 )) =Σ_(n=1) ^∞ ( (((2n)),(n) )/(4^n (2n+1)^2 ))+(1/4)Σ_(n=1) ^∞ (1/(4^(n−1) (2n+1)))−(1/4)Σ_(n=1) ^∞ (1/(4^(n−1) (2n+1)^2 )) =Σ_(n=1) ^∞ ( (((2n)),(n) )/(4^n (2n+1)^2 ))+(1/4)(Σ_(n=1) ^∞ (1/(4^(n−1) (2n+1)))−Σ_(n=1) ^∞ (1/(4^(n−1) (2n+1)^2 ))) =Σ_(n=1) ^∞ ( (((2n)),(n) )/(4^n (2^n +1)^2 ))+((π−1)/8) =(π^2 /4)−(π/4)+((π−1)/8) =(π^2 /8)−(π/8)−(1/8) =((π^2 −π−1)/8)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}+{n}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}^{{n}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\pi−\mathrm{1}}{\mathrm{8}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi}{\mathrm{4}}+\frac{\pi−\mathrm{1}}{\mathrm{8}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=\frac{\pi^{\mathrm{2}} −\pi−\mathrm{1}}{\mathrm{8}} \\ $$

Commented by Frix last updated on 01/Nov/24

Wrong. Your answer is ≈.716 but I get approximately ≈.123

$$\mathrm{Wrong}. \\ $$$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\approx.\mathrm{716}\:\mathrm{but}\:\mathrm{I}\:\mathrm{get}\:\mathrm{approximately} \\ $$$$\approx.\mathrm{123} \\ $$

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